Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

Given the equation and the solution , write the general solution.

I make the substitution and do all the proper work and get here:

and my brain gives out. I know this is easy, How do I find u?

February 18th 2011, 03:44 PM

TheEmptySet

Quote:

Originally Posted by VonNemo19

Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

Given the equation and the solution , write the general solution.

I make the substitution and do all the proper work and get here:

and my brain gives out. I know this is easy, How do I find u?

Well since divide both sides by it to get

now choose to get

February 18th 2011, 06:30 PM

Prove It

Of course, if you solve the Characteristic equation and realise you get a repeated root , then you immediately know that the solution of the DE is ...

February 18th 2011, 07:27 PM

TheEmptySet

Quote:

Originally Posted by Prove It

Of course, if you solve the Characteristic equation and realise you get a repeated root , then you immediately know that the solution of the DE is ...

Right but I think the point of the questions is to see how to get a 2nd linerly inpendant vector when you have a repeated root to the charateristic equation. Without this derivation how would we know what the 2nd vector is? Plus a great gereralization of this is variation of parameters!